Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{y - 2}{y^3 - 6y^2 + 8y} \div \dfrac{3y + 24}{3y^3 + 3y^2 - 168y} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{y - 2}{y^3 - 6y^2 + 8y} \times \dfrac{3y^3 + 3y^2 - 168y}{3y + 24} $ First factor out any common factors. $n = \dfrac{y - 2}{y(y^2 - 6y + 8)} \times \dfrac{3y(y^2 + y - 56)}{3(y + 8)} $ Then factor the quadratic expressions. $n = \dfrac {y - 2} {y(y - 2)(y - 4)} \times \dfrac {3y(y + 8)(y - 7)} {3(y + 8)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {(y - 2) \times 3y(y + 8)(y - 7) } { y(y - 2)(y - 4) \times 3(y + 8)} $ $n = \dfrac {3y(y + 8)(y - 7)(y - 2)} {3y(y - 2)(y - 4)(y + 8)} $ Notice that $(y - 2)$ and $(y + 8)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {3y(y + 8)(y - 7)\cancel{(y - 2)}} {3y\cancel{(y - 2)}(y - 4)(y + 8)} $ We are dividing by $y - 2$ , so $y - 2 \neq 0$ Therefore, $y \neq 2$ $n = \dfrac {3y\cancel{(y + 8)}(y - 7)\cancel{(y - 2)}} {3y\cancel{(y - 2)}(y - 4)\cancel{(y + 8)}} $ We are dividing by $y + 8$ , so $y + 8 \neq 0$ Therefore, $y \neq -8$ $n = \dfrac {3y(y - 7)} {3y(y - 4)} $ $ n = \dfrac{y - 7}{y - 4}; y \neq 2; y \neq -8 $